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Block Pulled Up Slope/Coefficient Of Kinetic Friction.

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Block A has a mass of 5.56 kg and is on a rough incline of 21.5 degrees to the horizontal. Block B has a mass of 4.66 kg and the coefficient of kinetic friction between Block A and the plane is 0.276. What is the acceleration of the blocks?

    2. Relevant equations
    I used a = m2g - m1gsin(theta) / (m1+m2)
    I see why thats wrong though, because I didn't use the kinetic friction.


    3. The attempt at a solution
    Again, I used the eq above to get
    a = (4.66)(9.81) - (5.56)(9.81)sin(21.5) / (5.56+4.66) = 2.52 m/s^2

    But I see why thats incorrect, I just don't know the correct method. haha.

    Thanks! :)
     
  2. jcsd
  3. Nov 2, 2008 #2

    Doc Al

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    Staff: Mentor

    I don't understand the setup. How are the two blocks related?
     
  4. Nov 2, 2008 #3
    just a sec, i'll draw a picture.

    [​IMG]

    Sorry that it took me a while to reply, I'm still working on the remaining questions.
     
  5. Nov 2, 2008 #4

    Doc Al

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    Staff: Mentor

    Well then, add the kinetic friction. What's the formula for finding the friction force? What's the normal force?
     
  6. Nov 2, 2008 #5
    I have N=f/mu but I don't know F or N. Hah. I'm usually not this terrible at physics, I promise.
     
  7. Nov 2, 2008 #6

    Doc Al

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    Staff: Mentor

    To find the normal force, analyze forces on the block perpendicular to the incline surface.
     
  8. Nov 2, 2008 #7
    See, thats the part that confuses me, the drawing of the arrows on the diagram. How do you know what direction N is pointing in? Whats mg? Etc.

    I'm really not understanding this entire topic.
     
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